1.4 Conditional probability

1 (a))

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{0.01 + 0.12 + 0.05}{ 0.01 + 0.02 + 0.05 + 0.11 + 0.06 + 0.08 + 0.13 } \\[0.5em] & = & \dfrac{0.18}{0.46} \\[0.5em] & = & 0.3913 \end{array} $$

(b)

$$ \begin{array}{r c l} P(C|A) & = & \dfrac{P(C \cap A)}{P(C)} \\[0.5em] & = & \dfrac{ 0.08 + 0.02 + 0.04 + 0.05 }{ 0.08 + 0.02 + 0.04 + 0.05 + 0.07 + 0.11 + 0.11 } \\[0.5em] & = & \dfrac{0.19}{0.48} \\[0.5em] & = & 0.3958 \end{array} $$

(c)

$$ \begin{array}{r c l} P(B|A \cap B) & = & \dfrac{P(B \cap (A \cap B))}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap A \cap B)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap B \cap A)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(B \cap A)}{P(A \cap B)} \\[0.5em] & = & \dfrac{P(A \cap B)}{P(A \cap B)} \\[0.5em] & = & 1 \end{array} $$

(d) Because

$$ \begin{array}{r c l} P(B|A \cup B) & = & \dfrac{P(B \cap (A \cup B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P((B \cap A) \cup (B \cap B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P((B \cap A) \cup (B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P((B \cap A) \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(B \cap A \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] \end{array} $$

$$ \begin{array}{c l} & P(B|A \cup B) \\[0.5em] = & \dfrac{P(B \cap (A \cup B))}{P(A \cup B)} \\[0.5em] = & \dfrac{P((B \cap A) \cup (B \cap B))}{P(A \cup B)} \\[0.5em] = & \dfrac{P((B \cap A) \cup (B))}{P(A \cup B)} \\[0.5em] = & \dfrac{P(B \cap A) - P(B) - P((B \cap A) \cap B)}{P(A \cup B)} \\[0.5em] = & \dfrac{P(B \cap A) - P(B) - P(B \cap A \cap B)}{P(A \cup B)} \\[0.5em] = & \dfrac{P(B \cap A) - P(B) - P(A \cap B \cap B)}{P(A \cup B)} \\[0.5em] = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] = & \dfrac{P(B \cap A) - P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] \end{array} $$

and $P(B \cap A)$ $=$ $0.01$ $+$ $0.02$ $+$ $0.05$ $=$ $0.08$, $P(A)$ $=$ $0.07$ $+$ $0.05$ $+$ $0.01$ $+$ $0.08$ $+$ $0.02$ $+$ $0.04$ $+$ $0.05$ $=$ $0.32$ and $P(B)$ $=$ $0.46$, then

$$ \begin{array}{r c l} \dfrac{P(B \cap A) + P(B) - P(A \cap B)}{P(A \cup B)} & = & \dfrac{0.08 + 0.46 - 0.08}{ \begin{array}{l} \phantom{+} 0.07 + 0.05 + 0.01 \\ + 0.08 + 0.02 + 0.04 \\ + 0.05 + 0.11 + 0.06 \\ + 0.08 + 0.13 \end{array} } \\[0.5em] & = & \dfrac{0.46}{0.7} \\[0.5em] & = & 0.6571. \end{array} $$

$$ \begin{array}{c l} & \dfrac{P(B \cap A) + P(B) - P(A \cap B)}{P(A \cup B)} \\[0.5em] = & \dfrac{0.08 + 0.46 - 0.08}{ \begin{array}{l} \phantom{+} 0.07 + 0.05 + 0.01 \\ + 0.08 + 0.02 + 0.04 \\ + 0.05 + 0.11 + 0.06 \\ + 0.08 + 0.13 \end{array} } \\[0.5em] = & \dfrac{0.46}{0.7} \\[0.5em] = & 0.6571. \end{array} $$

(e)

…

(f) Because

$$ \begin{array}{r c l} P(A \cap B|A \cup B) & = & \dfrac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} \\[0.5em] & = & \dfrac{P(A \cup B)}{P(A \cup B)} \\[0.5em] & = & 1, \end{array} $$

$$ \begin{array}{c l} & P(A \cap B|A \cup B) \\[0.5em] = & \dfrac{P((A \cap B) \cap (A \cup B))}{P(A \cup B)} \\[0.5em] = & \dfrac{P(A \cup B)}{P(A \cup B)} \\[0.5em] = & 1, \end{array} $$

because $(A \cap B)$ $\in$ $(A \cup B)$.

3 (a) Let $A$ denote event that card is ace of hearts and $B$ that the card is of red suit. Then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{1/52}{26/52} \\[0.5em] & = & \dfrac{1}{26} \\[0.5em] & \approx & 0.04 \end{array} $$

(b) If card is heart ($A$) and from red suit ($B$), then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{13/52}{26/52} \\[0.5em] & = & \dfrac{1/4}{1/2} \\[0.5em] & = & \dfrac{1}{8}. \end{array} $$

(c) If card is from red suit ($A$) and a heart ($B$), then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{1/26}{13/52} \\[0.5em] & = & \dfrac{1/26}{1/4} \\[0.5em] & \approx & 0.15 \end{array} $$

(d) If card is heart ($A$) and from black suit ($B$), then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{0}{26/52} \\[0.5em] & = & 0. \end{array} $$

Card can’t be heart if it’s from black suit.

(e) If card is king ($A$) and from red suit, then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{2/26}{26/52} \\[0.5em] & = & \dfrac{2}{13} \\[0.5em] & \approx & 0.15. \end{array} $$

(f) If card is king ($A$) and red picture card ($B$), then

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{1/12}{12/52} \\[0.5em] & \approx & 0.36. \end{array} $$

4 It the complement of $B$ is not empty and $A$ $\subset$ $B$, then $P(A)$ is smaller and a proper subset of $B$, or the same size as $P(A|B)$ and equal to $B$, but not bigger. An intuition why $A$ can’t be bigger is that, if it was, then

$$ \frac{P(A \cap B)}{P(B)} > 1 $$

which violates the axiom of probability where all events in a event space must be less than or equal to $1$.

5 The probability for ball being blue is $\frac{54}{150}$, so there are $150 - 54$ $=$ $96$ red balls in the bag. The probability that chosen red and shiny is then

$$ \dfrac{36}{96} = \dfrac{3}{8}. $$

The probability of a chosen ball being dull red is then $1 - \frac{3}{8}$ $=$ $\frac{5}{8}$, where $1$ is the event of ball being red.

6 Let $P(X)$ stand for the probability of event that car repair is on time and $P(Y|X)$ stand for the probability that the repair is satisfactory, given it’s on time. From the problem definition it’s known that $P(X)$ $=$ $0.77$ and $P(Y|X)$ $=$ $0.85$. The probability that repair is on time and is satisfactory is

$$ \begin{array}{c r c l} & P(X|Y) & = & \dfrac{P(Y \cap X)}{P(X)} \\[0.5em] \Longleftrightarrow & 0.85 & = & \dfrac{P(Y \cap X)}{0.77}. \end{array} $$

Multiplying both sides of the right hand equation shows that

$$ P(Y \cap X) = P(X \cap Y) = 0.85 \cdot 0.77 = 0.6545. $$

$$ \begin{array}{c l} & P(Y \cap X) \\[0.5em] = & P(X \cap Y) \\[0.5em] = & 0.85 \cdot 0.77 \\[0.5em] = & 0.6545. \end{array} $$

7 (a) If it’s raining today, the probability of raining tomorrow either increases, decreases or remains unchanged. (b) The probability that lottery winner has black hair, conditioned on that the lottery winner has brown eyes remains unchanged. (c) The probability that lottery winner has black hair, conditioned on that the lottery winner own a red car remains unchanged. (d) The probability that a lottery winner is more than $50$, conditioned on that the lottery winner is more than $30$ years old remains unchanged.

8 Since the assumption is that births are equally like on any day, the probability that a random person has a birthday on first day $P(A)$ $=$ $\frac{12}{365}$, because there are $12$ first days in a year, one for each month. When conditioned on March, i.e. what is the probability that the first person is born on first day given that day is in March is $\frac{1}{31}$, because there is only one first day on March. If the given month is February, the denominator of the previous fraction becomes $28$, so the probability for person born on first day given the month is February is $\frac{1}{28}$.

9 The events of batteries failing are:

$$ \begin{array}{| c c |} \hline E_1 & E_2 \\ (\text{I, II, III}) & (\text{I, III, II}) \\ \text{\small 0.11} & \text{\small 0.07} \\[1.0em] E_3 & E_4 \\ (\text{II, I, III}) & (\text{II, III, I}) \\ \text{\small 0.24} & \text{\small 0.39} \\[1.0em] E_5 & E_6 \\ (\text{III, I, II}) & (\text{III, II, I}) \\ \text{\small 0.16} & \text{\small 0.03} \\ \hline \end{array} $$

where a sequence $(A,B,C)$ denotes that battery of type $A$ fails first, then type $B$ and finally $C$. Given these the probabilities:

(a) Let probability of type $\text{I}$ battery lasting longest is $P(A)$. Then it not failing first is $P(A^\prime)$. Given these, $A$ conditional on $A^\prime$ is

$$ \begin{array}{r c l} P(A|A^\prime) & = & \dfrac{P(A \cap A^\prime)}{P(A^\prime)} \\[0.5em] & = & \dfrac{P(A \cap \{1 - P(A)\})}{1 - P(A)} \\[0.5em] & = & \dfrac{P(\{E_4,E_6\} \cap \{E_3,E_4,E_5,E_6\})}{\{E_3,E_4,E_5,E_6\}} \\[0.5em] & = & \dfrac{P(\{E_4,E_6\})}{P(\{E_3,E_4,E_5,E_6\})} \\[0.5em] & \approx & 0.51. \end{array} $$

$$ \begin{array}{c l} & P(A|A^\prime) \\[0.5em] = & \dfrac{P(A \cap A^\prime)}{P(A^\prime)} \\[0.5em] = & \dfrac{P(A \cap \{1 - P(A)\})}{1 - P(A)} \\[0.5em] = & \dfrac{P(\{E_4,E_6\} \cap \{E_3,E_4,E_5,E_6\})}{\{E_3,E_4,E_5,E_6\}} \\[0.5em] = & \dfrac{P(\{E_4,E_6\})}{P(\{E_3,E_4,E_5,E_6\})} \\[0.5em] \approx & 0.51. \end{array} $$

(b) The probability of type $\text{I}$ battery lasting longest ($A$) conditional on type $\text{II}$ failing first ($B$) is

$$ \begin{array}{r c l} P(A|B) & = & \dfrac{P(A \cap B)}{P(B)} \\[0.5em] & = & \dfrac{P(E_4)}{P(\{E_3,E_4\})} \\[0.5em] & \approx & 0.93. \end{array} $$

(c) The probability of type $\text{I}$ battery lasting longest ($A$) conditional on type $\text{II}$ lasting longest ($B$) is $0$ because $P(A \cap B)$ $=$ $0$ due to $A$ and $B$ being disjoint events.

(d) The probability of type $\text{I}$ battery lasting longest ($A$) is $P(A)$ $=$ $P(\{E_4,E_6\})$ and the probability of type $\text{II}$ battery not failing first ($B$) is $P(B)$ $=$ $P(\{E_1,E_2,E_5,E_6)$, so $A$ conditional on $B$ is $P(A \cap B)$ $/$ $P(B)$ $=$ $0.03$ $/$ $(0.11 + 0.07 + 0.16 + 0.03)$ $=$ $0.03$ $/$ $0.37$ $\approx$ $0.08$.

10 The given probability values for two assembly lines operations are,

$$ \begin{array}{ | c c c | } \hline (S,S) & (S,P) & (S,F) \\[-0.25em] \text{\small 0.02} & \text{\small 0.06} & \text{\small 0.05} \\[1em] (P,S) & (P,P) & (P,F) \\[-0.25em] \text{\small 0.07} & \text{\small 0.14} & \text{\small 0.20} \\[1em] (F,S) & (F,P) & (F,F) \\[-0.25em] \text{\small 0.06} & \text{\small 0.21} & \text{\small 0.19} \\[1em] \hline \end{array} $$

where $S$ stands for given line being shutdown, $P$ for line running at partial capacity and $F$ for line running at full capacity.

(a) The event that both lines are at full capacity is $A$ $=$ $\{ (F,F) \}$ and the event that neither of the lines are shutdown is $B$ $=$ $\{ (P,P), (P,F), (F,P), (F,F) \}$, so $A$ conditional on $B$ is $P(A|B)$ $=$ $P(A \cap B)$ $/$ $P(B)$ $=$ $0.19$ $/$ $(0.14$ $+$ $0.20$ $+$ $0.21$ $+$ $0.19)$ $\approx$ $0.26$.

(b) If event that at least one line is at full capacity $A$ $=$ $\{(S,F),$ $(P,F),$ $(F,S),$ $(F,P),$ $(F,F)\}$ and the event that neither line being shut is $B$ $=$ $\{(P,P),$ $(P,F),$ $(F,P),$ $(F,F)\}$, then $A$ conditional on $B$ is $P(A|B)$ $=$ $P(A \cap B)$ $/$ $P(B)$ $=$ $(0.20$ $+$ $0.21$ $+$ $0.19)$ $/$ $(0.14$ $+$ $0.20$ $+$ $0.21$ $+$ $0.19)$ $\approx$ $0.81$.

(c) When the event that at least one assembly line is running at full capacity is $A$ $=$ $\{(S,F),$ $(P,F),$ $(F,S),$ $(F,P),$ $(F,F)\}$ and the event where exactly one line is being shut down $B$ $=$ $\{(S,P),$ $(S,F),$ $(P,S),$ $(F,S)\}$, then $A$ conditional on $B$ is $P(A|B)$ $=$ $P(A \cap B)$ $/$ $P(B)$ $=$ $(0.05$ $+$ $0.06)$ $/$ $(0.06$ $+$ $0.05$ $+$ $0.07$ $+$ $0.06)$ $\approx$ $0.46$.

(d) Probability that neither line is at full capacity ($A$) conditional on at least one line is operating at partial capacity, that is, one line is either $P$ or $F$, or both, or in other words is not $(S,S)$ is $P(A|B)$ $=$ $P(A \cap B)$ $/$ $P(B)$ $=$ $(0.02$ $+$ $0.06$ $+$ $0.07$ $+$ $0.14)$ $/$ $(1 - 0.02)$ $\approx$ $0.30$.

…

12 Let gene being dominant or recessive be notated as $D$ or $R$, respectively, where gene either of type $A$ or $B$. When it’s stated that $P(D|B)$ $=$ $0.31$ and $P(B \cap D)$ $=$ $0.22$ $=$ $P(D \cap B)$, it can be said that $P(D|B)$ $=$ $P(D \cap B)$ $/$ $P(B)$ is equal to $0.31$ $=$ $0.22$ $/$ $P(B)$. Multiplying both sides with $P(B)$ and dividing with $0.31$, it can be seen that $P(B)$ $=$ $0.22$ $/$ $0.31$ $\approx$ $0.71$. Because of this $P(A)$ $=$ $1 - P(B)$ $\approx$ $0.29$, which is the probability of a gene being of type $A$.

13 AI Let the events $A$, $B$ and $C$, stand for component passing in performance, appearance and cost, respectively. Given probabilities are $P(B \cap C)$ $=$ $0.4$, $P(A \cap B \cap C)$ $=$ $0.31$, $P(A)$ $=$ $0.64$, $P(A^\complement \cap B^\complement \cap C^\complement)$ $=$ $0.19$ and $P(A^\complement \cap B \cap C^\complement)$ $=$ $0.06$.

(a) To solve the probability $P(A^\complement \cap B \cap C^\complement)$, that is, the probability that component fails performance and appearance, but passes on cost, the complement of $A$, i.e., component fails on performance needs to be found. Then, together with the given probabilities, the event space can be partitioned into four disjoint events:

$$ \begin{array}{l c c} P(A^\complement, B, C) & = & y \\ P(A^\complement, B, C^\complement) & = & 0.06 \\ P(A^\complement, B^\complement, C) & = & x \\ P(A^\complement, B^\complement, C^\complement) & = & 0.19 \end{array} $$

where $x$ is the probability under calculation. The probability $y$ $=$ $P(A^\complement, B, C)$ can be found by observing that it’s equal to $P(B \cap C)$ $-$ $P(A \cap B \cap C)$ $=$ $0.40$ $-$ $0.31$ $=$ $0.09$, so the above table of probabilities becomes:

$$ \begin{array}{l c c} P(A^\complement, B, C) & = & 0.09 \\ P(A^\complement, B, C^\complement) & = & 0.06 \\ P(A^\complement, B^\complement, C) & = & x \\ P(A^\complement, B^\complement, C^\complement) & = & 0.19 \end{array} $$

and so it can be said that $P(A^\complement)$ $=$ $P(A^\complement, B, C)$ $+$ $P(A^\complement, B, C^\complement)$ $+$ $P(A^\complement, B^\complement, C)$ $+$ $P(A^\complement, B^\complement, C^\complement)$ $=$ $0.09$ $+$ $0.06$ $+$ $x$ $+$ $0.19$. Because $P(A^\complement)$ $=$ $1$ $-$ $P(A)$ $=$ $0.36$, then $x$ $=$ $0.36$ $-$ $(0.09$ $+$ $0.06$ $+$ $0.19)$ $=$ $0.02$, so the probability that a component fails on performance and appearance, but succeeds on cost, that is,

$$ P(A^\complement, B^\complement, C) = 0.02. $$

(b) If a component passes on both appearance and cost, the probability that it passes on all three characteristics is $P(A$ $\cap$ $B$ $\cap$ $C$ $|$ $B$ $\cap$ $C)$ $=$ $P((A$ $\cap$ $B$ $\cap$ $C)$ $\cap$ $(B$ $\cap$ $C))$ $=$ $P(A$ $\cap$ $B$ $\cap$ $C$ $\cap$ $B$ $\cap$ $C)$ $/$ $P(B$ $\cap$ $C)$ $=$ $P($$(A \cap A)$ $\cap$ $(B \cap B)$ $\cap$ $C)$ $/$ $P(B$ $\cap$ $C)$ $=$ $P(A$ $\cap$ $B$ $\cap$ $C)$ $/$ $P(B$ $\cap$ $C)$ $=$ $0.775$.

14 When vegetable is good in taste ($T$), in size ($S$) and in appearance ($A$) and the given probabilities are $P(T)$ $=$ $0.78$, $P(T \cap S)$ $=$ $0.69$, $P(T \cap S^\complement \cap A)$ $=$ $0.5$ and $P(S \cup A)$ $=$ $P(S)$ $+$ $P(A)$ $-$ $P(S \cap P)$ $=$ $0.84$. (a) If the $P(S|T)$ $=$ $P(S \cap T)$ $/$ $P(T)$ $=$ $P(T \cap S)$ $/$ $P(T)$ $\approx$ $0.88$. (b) To find out the probability of vegetable being good taste, if a vegetable has bad size, that is, probability $P(T|(S^\complement \cap A^\complement))$

$$ \begin{array}{c l} = & \dfrac{P(T \cap (S^\complement \cap A^\complement))}{P(S^\complement \cap A^\complement)} \\[0.5em] = & \dfrac{P(T \cap S^\complement \cap A^\complement)}{P(S^\complement \cap A^\complement)} \\[0.5em] = & \dfrac{0.05}{P(S^\complement \cap A^\complement)} \end{array} $$

the denominator needs to be found. Noticing that $P(S^\complement \cap A^\complement)$ $=$ $P((S \cup A)^\complement)$ $=$ $1$ $-$ $P(S \cup A)$ $=$ $1$ $-$ $0.84$ $=$ $0.16$, the probability $P(T|(S^\complement \cap A^\complement)$ $=$ $0.05$ $/$ $0.16$ $=$ $0.3125$.

15 …

16 …

17 If the probabilities of expected revenue of a company is below expectation ($E_1$) with probability of $0.08$, slightly below expectation ($E_2$) with probability of $0.19$, meets the expectation ($E_3$) with probability of $0.26$, slightly above expectation ($E_4$) with probability of $0.36$, and above expectation ($E_5$) with probability of $0.11$, then if revenue is not below expectation, that is, revenue is expected or better ($\{E_3$ $\cap$ $E_4$ $\cap$ $E_5\}$), the probability of revenue being expected is

$$ \begin{array}{r c l} P(E_3 | \{E_3, E_4, E_5\}) & = & \dfrac{P(E_3 \cap \{E_3, E_4, E_5\})}{P(\{E_3, E_4, E_5\})} \\[0.5em] & = & \dfrac{P(E_3)}{P(E_3 \cap E_4 \cap E_5)} \\[1.0em] & \approx & 0.36. \end{array} $$

18 When the probabilities of an advertising campaign are: canceled before launch ($C$) is $0.10$ , launched and canceled early ($L_c$) is $0.18$, launched and runs it’s targeted time ($L_r$) is $0.43$ and launched and runs beyond extended length of it’s targeted time is ($L_e$) is $0.29$, the probability that it’s launched ($L$) is $P(L_c \cap L_r \cap L_e)$ $=$ $P(L)$ $=$ $0.18$ $+$ $0.43$ $+$ $0.29$ $=$ $0.90$. The probability that campaign runs it’s targeted time or beyond $T$, is $P(L_r \cap L_e)$ $=$ $P(T)$ $=$ $0.43$ $+$ $0.29$ $=$ $0.72$. Then, the probability that if advertising campaign is launched (i.e. $L$), that it runs it’s targeted time or beyond is $P(T|L)$ $=$ $P(T \cap L)$ $/$ $P(L)$, but because $T$ $\subset$ $L$, the probability $P(T|L)$ $=$ $0.72$ $/$ $0.90$ $=$ $0.80$.